Crossover frequency in RC filter circuits When the resistor's voltage equals the capacitor's voltage, crossover frequency is defined to be $1/(RC)$. I am aware that below crossover frequency, there will be a low pass area where only low frequencies pass through. Likewise, above crossover frequency, there will be a high pass area where only high frequencies pass through.
When the voltages equate each other, and you are at the crossover frequency, what frequencies pass through? Do the low and high frequencies cancel out and are blocked, or do both low and high frequencies transmit?
 A: The problem is that you are talking in absolutes: either a frequency is passed, or it is not.
An RC circuit does not work that way. No frequency is completely passed or blocked. With a C-R filter (R in series, then C to ground), the higher the frequency the better it is passed; with an R-C circuit it's the reverse: low frequencies are preferentially passed. Either way, the crossover frequency is the one that is attenuated by 3dB. 
Away from that frequency, the signal is attenuated more, or less. In the case of an R-C filter, only DC is passed unattenuated, and only infinite frequencies are completely eliminated. In the case of C-R it's the reverse: only infinite frequency is unattenuated, and DC is completely blocked. Any frequency in between is attenuated to some extent.
A: The most straightforward way to analyze reactive networks is with the concept of complex impedance. Analogous to resistance, the complex impedance tells us the relationship between voltage and current - but being a complex number, it allows us to take account of the phase difference "for free".
The impedance of a capacitor is $Z = \frac{1}{j\omega C}$ (we use $j$ rather than $i$ for the square root of -1 to prevent confusion with current, $I$.). Similarly for an inductor, $Z = j\omega L$. And inductor with series resistance would have $ZZ = j\omega L + R$.
A parallel LC network has a complex impedance that can be written as
$$Z = \frac{1}{\frac{1}{j\omega C}+j\omega L+R} = \frac{j\omega C}{1 - \omega^2 LC +j\omega RC}$$
As you can see, the denominator goes to zero at resonance in the case of a perfect inductor; if there is a bit of series resistance in the inductor, it won't quite go to zero...
You can now plot this as a function of frequency $\omega$ to tell you how much current can flow through your circuit as a function of frequency. For an inductor of 100 uH, a capacitor of 100 uF, and a series resistor varying between 0.5 Ohm and 5 Ohm, I get the following plot:

Which I generated with this Matlab code:
omega=linspace(0,10e3*2*pi, 1000);
L = 100e-6;
C = 100e-6;
figure
for R = [0.5 5]
    Z = 1j*omega*C./(1-omega.^2*L*C+1j*omega*R*C);
    plot(omega/(2*pi), abs(Z))
    hold on
end
title 'frequency response of LCR circuit'
xlabel 'frequency (Hz)'
ylabel '|Z| (Ohm)'
legend({'R=0.5', 'R=5'})

As you can see, the series resistor has a big impact on the answer.
And your question contains a paradox: "when the voltage is the same" implies that you are driving the filter with the resonant frequency - any other voltage being offered to the filter would not cause equal voltages on the two components...
A: Use complex impedance to understand this problem.


*

*The resistance is given by R

*The capacitive reactance is given by $\frac{-j}{\omega C}$, where $\omega=2\cdot \pi \cdot f$ 


In series, the components give the total impedance:
$$Z=R-\frac{j}{\omega C}$$
To analyse the impedance at the cross-over frequency ($f_0$, you will need to remember to  convert from real-imaginary, to real by taking the square root of the product of the complex impedance and it's conjugate
$$ \vert Z\vert =\sqrt{ Z Z^*}$$
Let's analyse the 'voltage' over the two components at $f_0$, by applying a source $V_S$.
$$V_R(f_0)=\frac{V_S\cdot R}{R-\frac{j}{2\cdot \pi \cdot f_0 \cdot C}}$$
$$V_C(f_0)=\frac{V_S\cdot R}{R-\frac{j}{2\cdot \pi \cdot f_0 \cdot C}}$$
$$R=\vert -\frac{j}{2\cdot \pi \cdot f_0 \cdot C}\vert$$
This graph, plotted with WXMAXIMA, shows how the 'voltage' at different frequencies are distributed over the two component. I simulated this with $ R = 10k\Omega$ and $C=10nF$ and $V_S=1$.
This gives, $f_{cut-off}=\frac{5000}{\pi}=1591.5Hz$
You mention low and high pass.  This is dependent upon which of the two components is being measured.  If you connect your oscilloscope to the resistor, and increase the frequency from zero to 10KHz, you will observe a high pass filter.  If you connect to the cap, you will observe a low-pass filter.

